Limit values of a variable in Julia JUMP

Question

How can I write a constraint that indicates that my variable can only take values that are multiples of 35?

Not sure if it makes sense, but you could try optimizing f(35 * x) instead of f(x) subject to x = 35 * k. In the first case, the value that goes into the function (35 * x) is always a multiple of 35, and you can simply optimize w.r.t. the x design variable which is now unconstrained. This can be implemented by multiplying the variable in question by 35 in your objective function. — ForceBru
– ForceBru, Commented Oct 26, 2022 at 19:43
Great idea, sadly it doesn't work for the problem i'm trying to solve :( — jvm.97
– jvm.97, Commented Oct 26, 2022 at 19:46
Does this answer your question? I need feasible solutions to be multiples of a number Julia — Sundar R
– Sundar R, Commented Oct 26, 2022 at 19:49
@jvm.97, posting the same question twice is frowned upon. You could also try asking on Julia's Discourse in the "Optimization" section: discourse.julialang.org/c/domain/opt/13 — ForceBru
– ForceBru, Commented Oct 26, 2022 at 20:10

Przemyslaw Szufel · Accepted Answer · 2022-10-26 20:22:36Z

1

using JuMP, Gurobi
m = Model(Gurobi.Optimizer)
@variable(m, x>=0, Int)
y = 35x

Now you can use y as any other variable in your optimization model.

If you want to have such constraint on the value of the objective (as the link that points to the deleted question) you could do (I reuse x from the previous example):

@variable(m,z1)
@variable(m,z2)

ob = 3z1 + 2z2

@constraint(m, ob == 35x)
@objective(m, Max, ob)

edited Oct 26, 2022 at 20:22

answered Oct 26, 2022 at 20:17

Przemyslaw Szufel

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2 Comments

jvm.97 Over a year ago

thanks! what if i want that constraint but for each variable? not the sum. I need every number of v[ j ] to be multiple of 35.. The problem i keep having is that v[j] is a Variable Ref

Przemyslaw Szufel Over a year ago

Having N the size of your variable you could just do @variable(m, x[1:N]>=0, Int); y=35 .* x

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